Question

Every rational \(x\) can be written in the form \(x=m / n\), where \(n>0\), and \(m\) and \(n\) are integers without any common divisors. When \(x=0\), we take \(n=1\). Consider the function \(f\) defined on \(R^{1}\) by $$f(x)=\left\\{\begin{array}{ll} 0 & (x \text { irrational }) \\ \frac{1}{n} & \left(x=\frac{m}{n}\right) \end{array}\right.$$ Prove that \(f\) is continuous at every irrational point, and that \(f\) has a simple discontinuity at every rational point.

Short answer

Short Answer: The function \(f(x)\) is continuous at every irrational point because the left and right limits, as well as the function value, are all equal to 0 for irrational points. On the other hand, the function has a simple discontinuity at every rational point because although the left and right limits exist and are equal to 0, the function value is not 0 (it is \(\frac{1}{n}\) for a rational point \(c = \frac{m}{n}\)).

Step-by-step solution

Recall the definition of continuity

A function \(f(x)\) is continuous at a point \(c\) if the following three conditions are satisfied: 1. \(f(c)\) is defined 2. \(lim_{x \to c^-} f(x) = L\) exists 3. \(lim_{x \to c^+} f(x) = L\) exists and \(f(c) = L\) We will apply this definition to our function \(f(x)\) for irrational and rational points.

Continuity at irrational points

Let \(c\) be an irrational point. We have to show that the given function \(f(x)\) is continuous at this point. That is, the three conditions for continuity are satisfied. 1. \(f(c)\) is defined: Since \(c\) is irrational, by definition, \(f(c) = 0\). 2. \(lim_{x \to c^-} f(x)\) exists: To evaluate the limit, we consider a sequence of rational and irrational numbers that approach \(c\) from the left. Let \(x_n = \frac{m}{n}\) and \(y_n\) be a sequence of rational and irrational numbers respectively, such that \(x_n < c\) for all \(n\). We need to show that \(lim_{n \to \infty} f(x_n) = lim_{n \to \infty} f(y_n) = 0\). Since \(x_n\) is rational, \(f(x_n) = \frac{1}{n}\), and \(f(y_n) = 0\) as \(y_n\) is irrational. As \(n \to \infty\), we have \(lim_{n \to \infty} f(x_n) = lim_{n \to \infty} \frac{1}{n} = 0\) and \(lim_{n \to \infty} f(y_n) = 0\). Thus, the left limit \(lim_{x \to c^-} f(x)\) exists and is equal to 0. 3. \(lim_{x \to c^+} f(x)\) exists: Similar to the previous case, we consider a sequence of rational and irrational numbers that approach \(c\) from the right. Let \(x_n = \frac{m}{n}\) and \(y_n\) be a sequence of rational and irrational numbers respectively, such that \(x_n > c\) for all \(n\). We need to show that \(lim_{n \to \infty} f(x_n) = lim_{n \to \infty} f(y_n) = 0\). Since \(x_n\) is rational, \(f(x_n) = \frac{1}{n}\), and \(f(y_n) = 0\) as \(y_n\) is irrational. As \(n \to \infty\), we have \(lim_{n \to \infty} f(x_n) = lim_{n \to \infty} \frac{1}{n} = 0\) and \(lim_{n \to \infty} f(y_n) = 0\). Thus, the right limit \(lim_{x \to c^+} f(x)\) exists and is equal to 0. Since all three conditions are satisfied, \(f(x)\) is continuous at every irrational point.

Simple discontinuity at rational points

Let \(c\) be a rational point of the form \(c = \frac{m}{n}\). We need to show that \(f(x)\) has a simple discontinuity at this point. A simple discontinuity occurs when both left and right limits exist and are finite but are not equal to the value of the function at the point. 1. \(f(c)\) is defined: Since \(c\) is rational, by definition, \(f(c) = \frac{1}{n}\). 2. Left limit \(lim_{x \to c^-} f(x)\) exists: Similar to the previous case, we consider a sequence of rational and irrational numbers that approach \(c\) from the left. Let \(x_k = \frac{p}{q}\) and \(y_k\) be a sequence of rational and irrational numbers respectively, such that \(x_k < c\). We need to show that \(lim_{k \to \infty} f(x_k) = lim_{k \to \infty} f(y_k) = 0\). Since \(x_k\) is rational, \(f(x_k) = \frac{1}{q}\), and \(f(y_k) = 0\) as \(y_k\) is irrational. As \(k \to \infty\), we have \(lim_{k \to \infty} f(x_k) = lim_{k \to \infty} \frac{1}{q} = 0\) and \(lim_{k \to \infty} f(y_k) = 0\). Thus, the left limit \(lim_{x \to c^-} f(x)\) exists and is equal to 0. 3. Right limit \(lim_{x \to c^+} f(x)\) exists: Similar to the previous case, we consider a sequence of rational and irrational numbers that approach \(c\) from the right. Let \(x_k = \frac{p}{q}\) and \(y_k\) be a sequence of rational and irrational numbers respectively, such that \(x_k > c\). We need to show that \(lim_{k \to \infty} f(x_k) = lim_{k \to \infty} f(y_k) = 0\). Since \(x_k\) is rational, \(f(x_k) = \frac{1}{q}\), and \(f(y_k) = 0\) as \(y_k\) is irrational. As \(k \to \infty\), we have \(lim_{k \to \infty} f(x_k) = lim_{k \to \infty} \frac{1}{q} = 0\) and \(lim_{k \to \infty} f(y_k) = 0\). Thus, the right limit \(lim_{x \to c^+} f(x)\) exists and is equal to 0. 4. Discontinuity at \(c\): since both left and right limits exist, are finite, and equal to 0, but \(f(c) = \frac{1}{n} \neq 0\), we have a simple discontinuity at the rational point \(c\). Therefore, we have proven that function \(f(x)\) is continuous at every irrational point, and has a simple discontinuity at every rational point.

Key concepts

Rational numbers

Rational numbers are numbers that can be expressed as the quotient of two integers. This means any number that can be written in the form \( \frac{m}{n} \) is considered rational. Here, \( m \) and \( n \) are integers, and crucially, \( n \) is not zero. For example, \( \frac{1}{2}, \frac{-3}{4}, \text{and } 5 \) are all rational numbers. The integer \( 5 \) can be represented as \( \frac{5}{1} \).

• Rational numbers include both positive and negative numbers.
• Decimals that terminate or repeat are also considered rational numbers, as they can be converted into fractions.
• The set of rational numbers is dense, meaning there are infinitely many rational numbers between any two rational or irrational numbers.

For example, between numbers 2 and 3, you can have 2.5, \( \frac{9}{4} \), and many others. Understanding rational numbers is key when discussing the continuity and discontinuity of functions.

Irrational numbers

Irrational numbers are numbers that cannot be written as a simple fraction or quotient of two integers. They exist in contrast to rational numbers. For instance, numbers such as \(\sqrt{2}\), \(\pi\), and \(e\) are irrational because they cannot be neatly expressed as \( \frac{m}{n} \) where both \( m \) and \( n \) are integers.

• Irrational numbers have non-terminating, non-repeating decimal expansions.
• They play a crucial role in various fields of mathematics and science.
• Despite not being expressible as fractions, irrational numbers perfectly fill the real number line along with rational numbers.

When examining functions and their continuity, irrational numbers are significant because they often represent points where continuity is investigated, particularly for problems like the ones involving function \( f(x) \). Understanding irrational numbers helps students grasp why certain limits behave the way they do at these points.

Discontinuity

Discontinuity in a function is where the function is not continuous at a point. At such points, the value of the function might suddenly jump or become undefined. In the context of the function \( f(x) \), the function exhibits a unique kind of discontinuity called a simple discontinuity at rational points.

• Discontinuity occurs when the limit from the left is different from the limit from the right at a certain point, or it doesn't match the actual function value at that point.
• In our specific example, though both the left and right limits at rational points approach 0, the function value \( f(c) \) for rational \( c \) is \( \frac{1}{n} \), which does not match the limit value.
• This difference results in what is described as a simple (or jump) discontinuity.

Such discontinuities are key areas where functions may behave unexpectedly, and understanding them helps in analyzing where and why functions break continuity.